In the figure, O Is the centre of the circle and HP is parallel to KL. NR = NP, ∠OKH = 54° and ∠RPN = 58°. Find
- ∠LKS
- ∠KLN
(a)
OH = OK = Radius
∠HKO = ∠KHO = 54° (Isosceles triangle, OHB)
∠KOS
= 54° + 54°
= 108° (Exterior angle of a triangle)
OK = OS = Radius
∠OSK
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠LKS = 36° (Alternate angles, KL//HE)
(b)
NR = NP
∠NPR = ∠NRP = 58° (Isosceles triangle NPF)
∠PNR
= 180° - 58° - 58°
= 64°
∠SNL = ∠PNR = 64° (Vertically opposite angles)
∠KLN
= 180° - 64°
= 116° (Interior angles)
Answer(s): (a) 36°; (b) 116°